Find $\lim_{x\to 1}\left(\frac{m}{(1-x)^m}-\frac{n}{(1-x)^n}\right)$ without using derivatives, the L'Hopital's rule or Taylor's series

Question

I need to solve this limit without using derivatives, the L'Hopital's rule or Taylor series: $$ \lim_{x\to 1}\left(\frac{m}{(1-x)^m}-\frac{n}{(1-x)^n}\right), \quad m,n\in\mathbb N. $$ I know that $$ \frac{m}{(1-x)^m}= \frac{m}{(1-x)(1+x+x^2+\ldots+x^{m-1})}, $$ $$ \frac{n}{(1-x)^n}= \frac{n}{(1-x)(1+x+x^2+\ldots+x^{n-1})}, $$ $$ \frac{m}{(1-x)^m}-\frac{n}{(1-x)^n}= \frac{m(1+x+x^2+\ldots+x^{n-1})-n(1+x+x^2+\ldots+x^{m-1})}{(1-x)(1+x+x^2+\ldots+x^{m-1})(1+x+x^2+\ldots+x^{n-1})}, $$ but I don’t know what to do next.

Answer 1

Let $E$ denote the expression.

$(1)$ If $m=n$ then clearly the asked limit is zero.

$(2)$ $m$ and $n$ even with $m\gt n$

$E=\dfrac{1}{(1-x)^n}\left(\dfrac{m}{(1-x)^{m-n}}-n\right)\to\infty(\infty-n)=\infty\times\infty=\infty$

$(3)$ $m$ and $n$ even with $m\lt n$.

Similarly to $(2)$ but we have now $E$ tends to $-\infty$.

Analogously if $m$ is even and $n$ is odd one has limit $\infty$ when $m\gt n$ and $-\infty$ when $m\lt n$ and if $m$ is odd and $n$ even the limit is $-\infty$ when $m\lt n$ and $\infty$ when $m\gt n$.

(The proof in detail is leave for the O.P. and you can follow $(2)$ for it)